In a metre-bridge,when a resistance in the left gap is $2 \ \Omega$ and an unknown resistance is in the right gap,the balance length is found to be $40 \ cm$. On shunting the unknown resistance with $2 \ \Omega$,the balance length changes by: (in $cm$)

When a wire is connected in the left gap of a metre bridge,the balancing point is at $40 \ cm$ from the left end of the bridge wire. If the wire in the left gap is stretched so that its length is doubled and again connected in the same gap,then the balancing point from the left end of the bridge wire is

In the figure shown,for given values of $R_1$ and $R_2$,the balance point for the jockey is at $40\,cm$ from $A$. When $R_2$ is shunted by a resistance of $10\, \Omega$,the balance point shifts to $50\,cm$. Find $R_1$ and $R_2$. $(AB = 1\,m)$

On interchanging the resistances,the balance point of a meter bridge shifts to the left by $10 \, cm$. The resistance of their series combination is $1 \, k\Omega$. How much was the resistance on the left slot before interchanging the resistances? ................... $\Omega$

In a meter bridge,a $30 \Omega$ resistance is connected in the left gap and a pair of resistances $P$ and $Q$ in the right gap. Measured from the left,the balance point is $37.5 \text{ cm}$ when $P$ and $Q$ are in series,and $71.4 \text{ cm}$ when they are in parallel. The values of $P$ and $Q$ (in $\Omega$) are:

When a metal conductor connected to the left gap of a meter bridge is heated,the balancing point